Core Pure: Complex Numbers Depth
In Core Pure, part of A-level Further Maths, complex numbers depth goes beyond the modulus-argument form and De Moivre's theorem to two further applications. First, writing z = cos theta + i sin theta gives z + 1/z = 2cos theta and z - 1/z = 2i sin theta, and De Moivre's theorem gives z^n + 1/z^n = 2cos(n theta) and z^n - 1/z^n = 2i sin(n theta); combining these with the binomial expansion lets a student express cos(n theta) in terms of powers of cos theta, or express powers such as cos^n(theta) and sin^n(theta) as a sum of cosines or sines of multiple angles. Second, combining loci on the Argand diagram into a single region, and finding the greatest or least value of |z| or arg(z) on that region, tests the geometry of circles and half-lines rather than just sketching them.
Method
- Write z = cos theta + i sin theta, so that z + 1/z = 2cos theta and z - 1/z = 2i sin theta, and by De Moivre's theorem z^n + 1/z^n = 2cos(n theta) and z^n - 1/z^n = 2i sin(n theta).
- To express cos^n(theta) or sin^n(theta) in terms of multiple angles, raise (z + 1/z)^n = (2cos theta)^n or (z - 1/z)^n = (2i sin theta)^n, expand with the binomial theorem, then group the terms in pairs z^k + 1/z^k (or z^k - 1/z^k) and convert each pair back to a cosine or sine of a multiple angle.
- To express cos(n theta) or sin(n theta) in terms of powers of cos theta and sin theta, apply De Moivre's theorem to (cos theta + i sin theta)^n, expand with the binomial theorem using i^2 = -1, then equate the real part to cos(n theta) or the imaginary part to sin(n theta), using sin^2(theta) = 1 - cos^2(theta) to leave a single variable where needed.
- For a locus or region question, sketch each condition separately on the Argand diagram (a circle |z - a - bi| = r, a half-line arg(z - a - bi) = k, or a perpendicular bisector |z - a| = |z - b|), then identify the part of the diagram that satisfies every condition at once.
- To find the greatest or least value of |z| on a circle |z - a - bi| = r, find the distance d = sqrt(a^2 + b^2) from the origin to the centre; if the origin lies outside the circle (d > r), the greatest value is d + r and the least value is d - r.
- To find the greatest value of arg(z) on a circle centred on the positive real axis, use the tangent line from the origin to the circle: if the centre is at distance d from the origin and the radius is r, the angle between the tangent line and the positive real axis is arcsin(r/d).
- Verify a multiple-angle identity by substituting a specific value of theta into both sides and checking the two sides agree numerically.
Worked example
Show that cos(5 theta) = 16cos^5(theta) - 20cos^3(theta) + 5cos(theta).
- Let c = cos theta and s = sin theta, so z = c + is. By De Moivre's theorem, z^5 = cos(5theta) + i sin(5theta).
- Expand z^5 = (c + is)^5 by the binomial theorem, then use i^2 = -1, i^3 = -i, i^4 = 1, i^5 = i to separate real and imaginary parts: z^5 = (c^5 - 10c^3s^2 + 5cs^4) + i(5c^4s - 10c^2s^3 + s^5).
- Equate the real part of z^5 to cos(5theta): cos(5theta) = c^5 - 10c^3s^2 + 5cs^4.
- Replace every s^2 with 1 - c^2, since sin^2(theta) = 1 - cos^2(theta), so the right-hand side is written purely in terms of c.
- Expand: c^5 - 10c^3(1 - c^2) + 5c(1 - c^2)^2 = c^5 - 10c^3 + 10c^5 + 5c(1 - 2c^2 + c^4) = c^5 - 10c^3 + 10c^5 + 5c - 10c^3 + 5c^5.
- Collect like terms and substitute back c = cos theta: cos(5theta) = 16cos^5(theta) - 20cos^3(theta) + 5cos(theta), as required.
Practice questions
Try each question, then tap to reveal the answer.
Q1Given that z = cos theta + i sin theta, write down the value of z + 1/z.Show answer
Answer: 2cos theta, since 1/z is the complex conjugate of z when |z| = 1, so z + 1/z = 2 Re(z) = 2cos theta.
Q2Given that z = cos theta + i sin theta, write down the value of z - 1/z.Show answer
Answer: 2i sin theta, since z - 1/z = (cos theta + i sin theta) - (cos theta - i sin theta) = 2i sin theta.
Q3Expand (z + 1/z)^3 as a sum of powers of z, using the binomial theorem.Show answer
Answer: z^3 + 3z + 3/z + 1/z^3 (from z^3 + 3z^2(1/z) + 3z(1/z)^2 + 1/z^3).
Q4Show that cos^3(theta) = (1/4)cos(3theta) + (3/4)cos(theta).Show answer
Answer: (2cos theta)^3 = (z+1/z)^3 = (z^3+1/z^3) + 3(z+1/z) = 2cos(3theta) + 6cos(theta), so 8cos^3(theta) = 2cos(3theta) + 6cos(theta), giving cos^3(theta) = (1/4)cos(3theta) + (3/4)cos(theta).
Q5The complex number z satisfies |z - 5 - 12i| = 4. Find the greatest and least possible value of |z|.Show answer
Answer: Distance from origin to centre (5,12) is sqrt(5^2+12^2) = sqrt(169) = 13. Greatest |z| = 13 + 4 = 17; least |z| = 13 - 4 = 9.
Q6The complex number z satisfies |z - 8| = 6. Find the least possible value of |z|.Show answer
Answer: Distance from origin to centre (8,0) is 8. Since 8 > 6, the least value of |z| is 8 - 6 = 2.
Q7The complex number z satisfies |z - 6| = 3. Find the greatest value of arg(z), giving your answer in radians to 3 significant figures.Show answer
Answer: The tangent line from the origin to the circle (centre (6,0), radius 3) makes an angle arcsin(3/6) = arcsin(0.5) = pi/6 with the positive real axis, so the greatest value of arg(z) is pi/6 = 0.524 radians (3 sf).
Q8Explain why the least possible value of |z| for a point z on the circle |z - a - bi| = r, where sqrt(a^2+b^2) > r, is sqrt(a^2+b^2) - r rather than r - sqrt(a^2+b^2).Show answer
Answer: A modulus cannot be negative. Since the origin lies outside the circle, the closest point on the circle to the origin is reached by moving from the centre towards the origin by the radius r, giving the positive value sqrt(a^2+b^2) - r; the alternative r - sqrt(a^2+b^2) would be negative here, which is impossible for |z|.
Exam-style questions
Written in the style of a A Level Further Maths exam paper, with a full mark scheme.
Show that sin(3theta) = 3sin(theta) - 4sin^3(theta).
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The complex number z satisfies |z - 3 - 4i| = 5. (a) Show that the locus of z passes through the origin. (b) Find the greatest value of |z|.
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Show that sin^4(theta) = (1/8)cos(4theta) - (1/2)cos(2theta) + 3/8.
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